anonymous
  • anonymous
Find all distinct roots(real or complex) of z^2 + (9+3i)z + (24+11i) in the form a+bi. Hint: You may have to complete the square
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
http://answers.yahoo.com/question/index?qid=20121124223053AAchfPc the bottom explanation of this link might be it, but i dont understand it
anonymous
  • anonymous
have you figured anything out?
sirm3d
  • sirm3d
\[\large z^2+(9+3i)z+(24-11i)=0\]\[\large z^2+(9+3i)z=-(24-11i)\]to complete squares, add (b/2)^2 to both sides, where b=9+3i \[\large z^2+(9+3i)z+\left( \frac{ 9+3i }{ 2 } \right)^2=-(24-11i)+\left( \frac{ 9+3i }{ 2 } \right)^2\]

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anonymous
  • anonymous
yeah im at this step... what do you do next?
anonymous
  • anonymous
\[(z + (\frac{ 9 + 3i }{ 2 }))^2 = -24 + 11i + 18 +\frac{ 27i }{ 2 }\]
anonymous
  • anonymous
\[z + (\frac{ 9+3i }{ 2 }) = \pm \sqrt{\frac{ -12 + 5i}{ 2 }} \]
anonymous
  • anonymous
i dunno, lol
anonymous
  • anonymous
i dont get how the person completed the square in the yahoo link..
anonymous
  • anonymous
\[z = -(\frac{ 9+3i }{ 2}) \pm \sqrt{\frac{ -12 + 5i }{ 2 }}\]
anonymous
  • anonymous
it would be so much easier if the square root wasnt there...we could just add them up easily
anonymous
  • anonymous
from the yahoo answers link: z² + (2−4i)z = (11+10i) z² + (2−4i)z + (-3 -4i) = (11+10i) + (-3 -4i)
anonymous
  • anonymous
what i dont get is where the (-3-4i) comes from

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